SOLUTION: can you help me understand how to find the vertex, the line of symmetry, and range with this problem for the graph of the function f(x)=-x^2+6x-8 a. find the vertex b. the

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: can you help me understand how to find the vertex, the line of symmetry, and range with this problem for the graph of the function f(x)=-x^2+6x-8 a. find the vertex b. the      Log On


   

Question 77060: can you help me understand how to find the vertex, the line of symmetry, and range with this problem
for the graph of the function f(x)=-x^2+6x-8
a. find the vertex
b. the line of symmetry
c. find the range
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a)
First factor out the negative 1
y=-1%28x%5E2-6x%2B8%29
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=1+x%5E2-6+x%2B8 Start with the given equation



y-8=1+x%5E2-6+x Subtract 8 from both sides



y-8=1%28x%5E2-6x%29 Factor out the leading coefficient 1



Take half of the x coefficient -6 to get -3 (ie %281%2F2%29%28-6%29=-3).


Now square -3 to get 9 (ie %28-3%29%5E2=%28-3%29%28-3%29=9)





y-8=1%28x%5E2-6x%2B9-9%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 9 does not change the equation




y-8=1%28%28x-3%29%5E2-9%29 Now factor x%5E2-6x%2B9 to get %28x-3%29%5E2



y-8=1%28x-3%29%5E2-1%289%29 Distribute



y-8=1%28x-3%29%5E2-9 Multiply



y=1%28x-3%29%5E2-9%2B8 Now add 8 to both sides to isolate y



y=1%28x-3%29%5E2-1 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=3, and k=-1. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2-6x%2B8 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2-6x%2B8%29 Graph of y=1x%5E2-6x%2B8. Notice how the vertex is (3,-1).



Notice if we graph the final equation y=1%28x-3%29%5E2-1 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x-3%29%5E2-1%29 Graph of y=1%28x-3%29%5E2-1. Notice how the vertex is also (3,-1).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.





So if we multiply everything by -1 we get
y=-%28x-3%29%5E2%2B1
So the vertex is (3,1)

b)
The line of symmetry will go through the vertex so the equation is
x=3

c)
The range is any value y can be, so by our graph we can see the range is
(, 1]
Here's our graph
+graph%28+300%2C+200%2C+-6%2C+5%2C+-10%2C+10%2C+-x%5E2%2B6x-8%29+ graph of y=-%28x-3%29%5E2%2B1